Does Cox Regression have an underlying Poisson distribution?

Yes, there is a link between these two regression models. Here is an illustration:

Suppose the baseline hazard is constant over time: h0(t)=λ$h_0(t)=\lambda$. In that case, the survival function is

S(t)=exp(−∫t0λdu)=exp(−λt)$S(t)=\exp (-{\int }_0^t\lambda du)=\exp (-\lambda t)$

and the density function is

f(t)=h(t)S(t)=λexp(−λt)$f(t)=h(t)S(t)=\lambda \exp (-\lambda t)$

This is the pdf of an exponential random variable with expectation λ−1${\lambda }^{-1}$.

Such a configuration yields the following parametric Cox model (with obvious notations):

hi(t)=λexp(x′iβ)$h_i(t)=\lambda \exp (x_i^{\prime }\beta )$

In the parametric setting the parameters are estimated using the classical likelihood method. The log-likelihood is given by

l=∑i{dilog(hi(ti))−tihi(ti)}$l=\sum _i\{d_i\log (h_i(t_i))-t_i{h}_i(t_i)\}$

where di$d_i$is the event indicator.

Up to an additive constant, this is nothing but the same expression as the log-likelihood of the di$d_i$'s seen as realizations of a Poisson variable with mean μi=tihi(t)${\mu }_i=t_i{h}_i(t)$.

As a consequence, one can obtain estimates using the following Poisson model:

log(μi)=log(ti)+β0+x′iβ$\log ({\mu }_i)=\log (t_i)+{\beta }_0+x_i^{\prime }\beta$

where β0=log(λ)${\beta }_0=\log (\lambda )$.